Question

Consider a water pipe of length \(L=17 \mathrm{~m}\), inner radius $r_{1}=15 \mathrm{~cm}\(, outer radius \)r_{2}=20 \mathrm{~cm}$, and thermal conductivity \(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is generated in the pipe material uniformly by a \(25-\mathrm{kW}\) electric resistance heater. The inner and outer surfaces of the pipe are at \(T_{1}=60^{\circ} \mathrm{C}\) and \(T_{2}=80^{\circ} \mathrm{C}\), respectively. Obtain a general relation for temperature distribution inside the pipe under steady conditions and determine the temperature at the center plane of the pipe.

Short answer

Answer: The temperature at the center plane of the pipe is approximately 64.92°C.

Step-by-step solution

1. Obtain the expression for heat transfer in the radial direction

For a cylindrical pipe, the heat transfer per unit length, q, in the radial direction can be written as: \[q = k A \frac{dT}{dr}\] where - q is the heat transfer per unit length in the radial direction (W/m), - k is the thermal conductivity of the material (W/m*K), - A is the cross-sectional area for heat transfer, and - dT/dr is the temperature gradient in the radial direction. For a cylindrical pipe, the cross-sectional area A is given by: \[A = 2\pi rL\]

2. Integrate the heat transfer equation

Substitute the cross-sectional area A into the heat transfer equation, and integrate the equation over the radial direction to obtain the general relation for temperature distribution inside the pipe: \[q = k(2\pi rL) \frac{dT}{dr}\] Separating variables and integrating the equation, we get: \[\frac{q}{2\pi kL}\int_{r_1}^{r} \frac{dr'}{r'} = \int_{T_1}^{T} dT'\]

3. Evaluate the integrals

Evaluating the integrals and solving for the temperature T as a function of the radial position r, we obtain the general relation for temperature distribution inside the pipe: \[T(r) = T_1 + \frac{q}{2\pi kL} \ln \left( \frac{r}{r_1} \right)\]

4. Calculate the heat generated per unit length

Since heat is generated uniformly by a 25 kW electric resistance heater, the heat generated per unit length can be calculated as: \[q = \frac{P}{L} = \frac{25,000 \text{ W}}{17 \text{ m}} = 1470.588 \text{ W/m}\]

5. Determine the temperature at the center plane of the pipe

To determine the temperature at the center plane of the pipe, we use the average of the radii \(r_1\) and \(r_2\): \[r_m = \frac{r_1 + r_2}{2} = \frac{0.15 \text{ m} + 0.20 \text{ m}}{2} = 0.175 \text{ m}\] Substitute the parameters \(q\), \(r_1\), \(k\), \(L\), and \(T_1\) into the general relation for the temperature distribution, and calculate \(T(r_m)\) using \(r_m\): \[T(0.175) = 60 + \frac{1470.588}{2\pi (14)(17)}\ln\left(\frac{0.175}{0.15}\right) = 64.92^{\circ}\text{C}\] Thus, the temperature at the center plane of the pipe is approximately \(64.92^{\circ} \mathrm{C}\).